What is the index of $\mathbb{Z} \times2\mathbb{Z}$ in $\mathbb{Z} \times\mathbb{Z}$ As far as I know
$$\mathbb{Z}\times2\mathbb{Z}=\{(m,2n);m, n\in\mathbb{Z} \}, \ \ \mathbb{Z}\times\mathbb{Z}=\{(p,q);p, q\in\mathbb{Z} \}. $$
In order to computer the cosets, do I have to consider the sum
$$(m+p, 2n+q)$$
And analise the cases where $m, n, p, q$ are odd and/or even? For example, if all of them are even, then
$$(p, q) +\mathbb{Z}\times\mathbb{Z}=\{(\overline{0},\overline{0})\}, $$
since $m+p$ and $2n+q$ are even and, hence, they belong to $\overline {0} $.
Is this the way?
 A: By definition, two elements $(u,v), (x,y) \in\mathbb Z\times \mathbb Z$ are in the same coset iff $(u,v)-(x,y)\in \mathbb Z\times 2\mathbb Z$.
Let $(a,b) \in \mathbb Z\times \mathbb Z$. There are two possibilites: $b$ is either even or odd.
If $b$ is even, then $(a,b)-(0,0)\in \mathbb Z\times 2\mathbb Z$.
If $b$ is odd,  then $(a,b)-(0,1)\in \mathbb Z\times 2\mathbb Z$, whence
there are two cosets: $(0,0)+\mathbb Z\times 2\mathbb Z$ and $(0,1)+\mathbb Z\times 2\mathbb Z$, i.e. the index is 2.
A: $\newcommand{\ZZ}{\mathbb{Z}}$
Firstly, note that $\ZZ\times\ZZ$ is the bigger group. Using your notation, you want to count the number of cosets $$(p,q)+\ZZ\times2\ZZ$$ where $p,q\in\ZZ$. What happens to $(p,0)$? Well it goes to the coset $(0,0)+\ZZ\times2\ZZ$ because $p\in\ZZ$. What about $(p,2q')$ where $q'\in\ZZ$? This also is the identity in the quotient. And if the second entry was odd, then this corresponds to $(0,1)+\ZZ\times2\ZZ$. Since these are all the cases, then the index is 2.
In general, let $G,H,K,J$ be groups such that $H\leq G$ and $J\leq K$. The index is $$[G\times K:H\times J]=[G:H][K:J]$$ by the properties of the direct product.
A: The index is $2$. The $2$ cosets are $\Bbb Z\oplus 2\Bbb Z$ and
$(0,1)+(\Bbb Z\oplus 2\Bbb Z)$.
